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Difficulty: MediumUnit Digit and Cyclicity

What is the unit digit of the expression K=(377123+428164)×533200−61986K = (377^{123} + 428^{164}) \times 533^{200} - 619^{86}?

Answer: 8

Answer

The unit digit of the expression is 8.
Using cyclicity principles for each power term, 377123377^{123} yields unit digit 3 (remainder 3), 428164428^{164} yields unit digit 6 (remainder 0 →\rightarrow 4th power), 533200533^{200} yields unit digit 1 (remainder 0 →\rightarrow 4th power), and 61986619^{86} yields unit digit 1 (even power of 9). Combining these gives (3+6)×1−1=8(3 + 6) \times 1 - 1 = 8.

Step-by-Step Solution

1
Find the unit digit of 377123377^{123}
Unit digit is 3
The unit digit of 377 is 7, which follows a cyclicity pattern of 4 (7, 9, 3, 1). Dividing the exponent 123 by 4 leaves a remainder of 3. Thus, the unit digit is given by 73=343→37^3 = 343 \rightarrow 3.
2
Find the unit digit of 428164428^{164}
Unit digit is 6
The unit digit of 428 is 8, which has a cyclicity of 4 (8, 4, 2, 6). The exponent 164 is divisible by 4 (remainder 0), which corresponds to the 4th position in the cyclicity cycle (84=4096→68^4 = 4096 \rightarrow 6).
3
Evaluate the sum inside the parentheses (377123+428164)(377^{123} + 428^{164})
Unit digit is 9
Adding the individual unit digits gives 3+6=93 + 6 = 9.
4
Find the unit digit of 533200533^{200}
Unit digit is 1
The unit digit of 533 is 3, which has a cyclicity of 4 (3, 9, 7, 1). Exponent 200 is divisible by 4 (remainder 0), corresponding to the 4th power in the cycle (34=81→13^4 = 81 \rightarrow 1).
5
Multiply the bracketed sum by 533200533^{200}
Unit digit is 9
Multiplying the unit digits yields 9×1=99 \times 1 = 9.
6
Find the unit digit of 61986619^{86}
Unit digit is 1
The unit digit of 619 is 9, which has a cyclicity of 2 (9 for odd powers, 1 for even powers). Since 86 is even, the unit digit is 1.
7
Subtract 61986619^{86} from the product
Unit digit is 8
Subtracting the unit digits gives 9−1=89 - 1 = 8.

Key Concept

Unit Digit and Cyclicity
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Topics

Unit Digit and Cyclicity

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